Problem: Is ${401814}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {401814}= &&{4}\cdot100000+ \\&&{0}\cdot10000+ \\&&{1}\cdot1000+ \\&&{8}\cdot100+ \\&&{1}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {401814}= &&{4}(99999+1)+ \\&&{0}(9999+1)+ \\&&{1}(999+1)+ \\&&{8}(99+1)+ \\&&{1}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {401814}= &&\gray{4\cdot99999}+ \\&&\gray{0\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {4}+{0}+{1}+{8}+{1}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${401814}$ is divisible by $3$ if ${ 4}+{0}+{1}+{8}+{1}+{4}$ is divisible by $3$ Add the digits of ${401814}$ $ {4}+{0}+{1}+{8}+{1}+{4} = {18} $ If ${18}$ is divisible by $3$ , then ${401814}$ must also be divisible by $3$ ${18}$ is divisible by $3$, therefore ${401814}$ must also be divisible by $3$.